\(f(x)=x^2\) is strictly increasing on \([0, \infty) \) i.e. \(x < y \implies f(x) < f(y) \)

 Prove: f(x) = \(x^2\) is strictly increasing on [0, \(\infty\)) i.e. x < y \(\rightarrow\)  f(x) < f(y)

Case 1: x = 0

Assume x < y. 

When x = 0, y > 0 we have \(x^2\) = 0 and \(y^2\) > 0

\(\therefore\) we have shown when x = 0, f(x) < f(y)

Case 2: x \(\neq\) 0

Assume x < y

Then multiply both sides of the inequality x < y by x and y

After multiplying both sides of the inequality we have: \(x^2\) < xy and yx < \(y^2\) 

Now if we combine these two inequalities together we have: \(x^2\) < xy < \(y^2\)

Using the transitive property: a < b and b < c \(\rightarrow\) a < c we get: \(x^2\) < \(y^2\)

\(\therefore\) we have shown when x \(\neq\) 0, f(x) < f(y)

In conclusion, we have shown if x < y \(\rightarrow\) f(x) < f(y)